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प्रश्न
Solve the following:
Given three identical boxes, I, II, and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
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उत्तर
Let event B1: Selecting box I have two gold coins,
event B2: Selecting box II having two silver coins,
event B3: Selecting box III having one silver and one gold coin,
event G: Coin is gold.
P(B1) = P(B2) = P(B3) = `1/3`
`"P"("G"/"B"_1) = 1, "P"("G"/"B"_2) = 0, "P"("G"/"B"_3) = 1/2`
P(G) = `"P"("B"_1) "P"("G"/"B"_1) + "P"("B"_2) "P"("G"/"B"_2) + "P"("B"_3) "P"("G"/"B"_3)`
= `1/3[1 + 0 + 1/2]`
= `1/3(3/2)`
= `1/2`
To find the probability that the other can in the box is also gold. Which is possible only when it is drawn from box I.
∴ Required probability = `"P"("B"_1/"G")`
By Bayes’ theorem,
`"P"("B"_1/"G") = ("P"("B"_1)"P"("G"/"B"_1))/("P"("G"))`
= `(1/3(1))/(1/2)`
= `2/3`
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Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.
Given that P(A) = `square`, P(C) = `square`
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`= square * square + square * square + square * square`
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= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
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